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02 Jul 2017, 02:23
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:23) correct
33%
(02:12)
wrong
based on 336
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A sum of money invested under simple interest, amounts to $1200 in three years and $1500 in five years. What is the rate at which the sum of money was invested?
(A) 10%
(B) 15%
(C) 20%
(D) 25%
(E) 45%
(A) 10%
(B) 15%
(C) 20%
(D) 25%
(E) 45%
02 Jul 2017, 03:44
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The amount at the end of 5 years is 1500 and at the end of 3 years is 1200
Total interest earned in these 2 years = 1500-1200 = 300
Interest earned per year = \(\frac{300}{2}\) = 150
The original amount invested at the beginning of year 1 = 1200-3*(150) = 1200-450 = 750
Annual rate of interest = \(\frac{150}{750}\) = \(\frac{1}{5}\) = 20%
Answer C
Total interest earned in these 2 years = 1500-1200 = 300
Interest earned per year = \(\frac{300}{2}\) = 150
The original amount invested at the beginning of year 1 = 1200-3*(150) = 1200-450 = 750
Annual rate of interest = \(\frac{150}{750}\) = \(\frac{1}{5}\) = 20%
Answer C
General Discussion
sashiim20
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Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 608
02 Jul 2017, 03:15
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Bunuel
Principal, Rate is same for \(3\) years and \(5\) years.
\(Amount = P + SI\) ------------ (\(SI =\) Simple interest. \(SI = \frac{Principle * Rate * Time}{100}\))
\(1200 = P + SI_3 => P = 1200 - SI_3\) --------- (i) ---------- (\(SI_3 = SI\) for \(3\) years)
\(1500 = P + SI_5 => P = 1500 - SI_5\) ---------- (ii) ---------- (\(SI_5 = SI\) for \(5\) years)
Equating (i) and (ii), we get;
\(1200 - SI_3 = 1500 - SI_5\)
\(1200 - \frac{PR*3}{100} = 1500 - \frac{PR*5}{100}\)
\(120000 - PR*3 = 150000 - PR*5\)
\(PR*5 - PR*3 = 150000 - 120000\)
\(PR(5 - 3) = 30000\)
\(PR = \frac{30000}{2} = 15000\)
Substituting value of \(PR\) in \(SI_3\), we get;
\(1200 = P + \frac{PR*3}{100}\)
\(1200 = P + \frac{15000*3}{100} = P + 450\)
\(P = 1200 - 45 = $750\)
\(SI\) for amount \(1200\) for \(3\) years \(= $450\)
\(450 = \frac{PR*3}{100}\)
\(R = \frac{450 * 100}{3*750} = 20\)%.
Answer (C)...
